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1127 - 咸鱼文章
Time Limit:1s Memory Limit:128MByte
Submissions:488Solved:200
elttiL moT nwod eht teerts sllac ruo god " ehT peek god " . piZ si a peehs god . tuB nehw moT seirt ot yas " peeS " , ti semoc tuo " peek " . dnA ni a yaw moT si thgir . piZ si syawla gnignirb sgniht oh rof su ot peek ! ll‘I llet uoy tuoba emos fo meht .
s‘piZ tsrif tneserp saw a eohs . tI saw edam fo neerg klis .
eW t‘ndid wonk woh piZ dnuof eht eohs . tuB retfa a tnemom yraM , ym gib retsis , dlot em eht eohs dah a egnarts llems . I deddon dna dleh ym eson . " tahW od uoy kniht ti si ? "
" tI sllems ekil gnihtemos rof gninaelc . I kniht enoemos deirt ot naelc a tops ffo eht eohs . nehT eh tup ti ta eht rood ot yrd . "
" gnolA emac piZ . dnA eyb-doog eohs ! " I dias . " eW dluohs ekat ti kcab . "
" eW t‘nac " . dias ym rettsis .
" ebyaM elttil moT si thgir , " yraM dias . " ebyaM piZ si a peek god ! "
你正在做英语阅读,可哪知这是一篇咸鱼文章,整个文章的所有单词都是翻转的,你很慌。
不过你是咸鱼程序员,你可以写代码将这篇文章翻转回来,那么翻转回来吧。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #define N 100005 5 using namespace std; 6 char s[N]; 7 char st[N]; 8 int main(){ 9 char c; 10 int a=0; 11 int k=0; 12 while((c=getchar())!=EOF){ 13 if((c>=‘A‘&&c<=‘Z‘)||(c>=‘a‘&&c<=‘z‘)){ 14 st[a++]=c; 15 }else{ 16 if(strlen(st)){ 17 for(int i=a-1;i>=0;i--){ 18 s[k++]=st[i]; 19 } 20 s[k++]=c; 21 a=0; 22 }else{ 23 s[k++]=c; 24 } 25 } 26 } 27 for(int i=0;i<k;i++){ 28 cout<<s[i]; 29 } 30 return 0; 31 }
1127 - 咸鱼文章